Java ArrayList同样的字符串分组并获取出现频率最高的字符串

huangapple go评论62阅读模式
英文:

Java ArrayList same string grouping and get most frequent String

问题

我有一个包含字符串的列表并且将相同的字符串分组

    List<String> allTypes = new ArrayList<String>();

    Map<String, Long> count = allTypes.stream()
                .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

然后我获取出现频率最高的字符串的计数

    Long max = Collections.max(count.values());

现在我不仅想要出现频率最高的字符串的计数还想要关联的字符串该列表是随机填充的其中的字符串来自另一个列表
英文:

i have a List with Strings and group the same Strings.

List&lt;String&gt; allTypes = new ArrayList&lt;String&gt;();

Map&lt;String, Long&gt; count = allTypes.stream()
			.collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

and then i get the count of the most frequent String

Long max = Collections.max(count.values());

Now i dont want only the count of the most frequent String i want the
associated String too. The List is randomly filled with Strings from a other List.

答案1

得分: 5

我认为你正在寻找:

使用Optional的版本:

Optional<Map.Entry<String, Long>> maxEntryByValue = count.entrySet()
        .stream()
        .max(Comparator.comparing(Map.Entry::getValue));

如果你想要不使用Optional,可以使用以下版本:

Map.Entry<String, Long> maxEntryByValue = count.entrySet()
        .stream()
        .max(Comparator.comparing(Map.Entry::getValue))
        .orElse(null); // 或任何默认值,或者你可以使用 orElseThrow(..)
英文:

I think you are looking to:

Optional&lt;Map.Entry&lt;String, Long&gt;&gt; maxEntryByValue = count.entrySet()
        .stream()
        .max(Comparator.comparing(Map.Entry::getValue));

Or if you want it without Optional, you can use:

Map.Entry&lt;String, Long&gt; maxEntryByValue = count.entrySet()
        .stream()
        .max(Comparator.comparing(Map.Entry::getValue))
        .orElse(null); // or any default value, or you can use orElseThrow(..)

答案2

得分: 0

你可以遍历这些条目,如果value与你的targetValue相匹配,就将该键存储在Set<K> keys中。这将包含所有具有特定targetValue的键。

    for (Map.Entry<String, Object> entry : map.entrySet()) {
      String key = entry.getKey();
      Object value = entry.getValue();
      if (value.equals(targetValue)) {
        keys.add(entry.getKey());//将键添加到set<K> keys中
      }
    }
    return keys; //所有具有相同targetValue的键将在此集合中
英文:

You can iterate over the entries and if the value matches your targetValue then store that key in the Set&lt;K&gt; keys. This will contain all the keys having that particular targetValue

    for (Map.Entry&lt;String, Object&gt; entry : map.entrySet()) {
      String key = entry.getKey();
      Object value = entry.getValue();
      if (value.equals(targetValue)) {
        keys.add(entry.getKey());//add the keys to the set&lt;K&gt; keys
      }
    }
    return keys; //all the keys having the same targetValue will be in this set

huangapple
  • 本文由 发表于 2020年10月15日 23:48:18
  • 转载请务必保留本文链接:https://go.coder-hub.com/64375273.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定